Question Number 139192 by EnterUsername last updated on 23/Apr/21 $$\mathrm{If}\:{z}_{\mathrm{1}\:} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{complex}\:{n}\mathrm{th}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{which}\:\mathrm{sub}- \\ $$$$\mathrm{tend}\:\mathrm{right}\:\mathrm{angle}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{then}\:{n}\:\mathrm{must}\:\mathrm{be}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4K}+\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{4K}+\mathrm{2} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{4K}+\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4K} \\ $$ Answered by mr W last…
Question Number 73649 by aliesam last updated on 14/Nov/19 $${find}\:{the}\:{range} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{6}−\sqrt{{x}+\mathrm{2}}} \\ $$ Commented by mathmax by abdo last updated on 14/Nov/19…
Question Number 139165 by mathlove last updated on 23/Apr/21 Answered by mr W last updated on 23/Apr/21 $$\mathrm{13}+\sqrt{\mathrm{48}}=\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}=\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}−\sqrt{\mathrm{13}+\sqrt{\mathrm{48}}}=\mathrm{5}−\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right)=\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{5}−\sqrt{\mathrm{13}+\sqrt{\mathrm{48}}}}=\mathrm{3}+\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{2}+\sqrt{\mathrm{3}} \\…
Question Number 73619 by L.Messi last updated on 14/Nov/19 Commented by MJS last updated on 14/Nov/19 $$\mathrm{what}'\mathrm{s}\:\mathrm{your}\:\mathrm{idea},\:\mathrm{Sir}\:\mathrm{L}.\:\mathrm{Messi}? \\ $$ Commented by L.Messi last updated on…
Question Number 8071 by marcusvsoliveira last updated on 29/Sep/16 $${Solve}\:−\mathrm{2}{x}\left({x}+\frac{\mathrm{7}}{\mathrm{2}}\right)−\left({x}−\mathrm{3}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$ Commented by FilupSmith last updated on 29/Sep/16 $$−\mathrm{2}{x}\left({x}+\frac{\mathrm{7}}{\mathrm{2}}\right)−\left({x}−\mathrm{3}\right)^{\mathrm{2}} =−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{7}{x}\right)−\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right) \\…
Question Number 139118 by EnterUsername last updated on 22/Apr/21 $$\mathrm{Let}\:{z}\:\mathrm{be}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}.\:\mathrm{If}\:\mid{z}+\mathrm{1}\mid=\mid{z}−\mathrm{1}\mid \\ $$$$\mathrm{and}\:\mathrm{arg}\left(\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\right)=\frac{\pi}{\mathrm{4}}.\:\mathrm{Then}\:{z}\:\mathrm{is}\:?\: \\ $$ Answered by qaz last updated on 22/Apr/21 $$\mid{z}+\mathrm{1}\mid=\mid{z}−\mathrm{1}\mid \\ $$$$\Rightarrow{z}={a}+{bi}={bi} \\…
Question Number 139116 by EnterUsername last updated on 22/Apr/21 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{z}\overline {\left(\mathrm{z}−\mathrm{2i}\right)}=\mathrm{2}\left(\mathrm{z}+\mathrm{i}\right)\:\mathrm{is}\:? \\ $$ Answered by qaz last updated on 22/Apr/21 $${z}\centerdot\overline {{z}−\mathrm{2}{i}}={z}\overset{−} {{z}}+{z}\overline…
Question Number 139113 by mathdanisur last updated on 22/Apr/21 $${if},\:{x}=\mathrm{3}+\sqrt{\mathrm{7}} \\ $$$${find},\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$ Answered by MJS_new last updated on 22/Apr/21 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{127}+\mathrm{48}\sqrt{\mathrm{7}}}}=\sqrt{\mathrm{127}−\mathrm{48}\sqrt{\mathrm{7}}} \\…
Question Number 73574 by A8;15: last updated on 13/Nov/19 Commented by malwaan last updated on 14/Nov/19 $$\mathrm{5}+\mathrm{40}+\mathrm{5}=\:\mathrm{50} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 139114 by EnterUsername last updated on 22/Apr/21 Commented by mr W last updated on 22/Apr/21 $${i}'{ll}\:{try}\:{to}\:{give}\:{a}\:{proof}\:{when}\:{i}\:{have}\:{time}. \\ $$ Commented by EnterUsername last updated…